Question

how do you find y"(x) for

y = the square root of 5x^2 + 4

the 5x^2 + 4 is all under the square root

Answer 1

chain rule ...

Answer 2

\[\frac d{dx}[f(x)]^n=n[f(x)]^{n-1}~*~\frac d{dx}f(x)\]

Answer 3

look like the second round will involve a product rule

Answer 4

But how do you get rid of the square root in order to do the first derivative

Answer 5

it helps to recall your algebra :)\[\sqrt{M}=M^{1/2}\]

Answer 6

so, is it (5x^2 + 4) ^1/2 ?

Answer 7

yes

Answer 8

\[(5x^2+4)^{1/2}\to 5x*(5x^2+4)^{-1/2}\]

Answer 9

therefore, it would be:
1/2 (5x^2 + 4 ) ^-1/2 ?

Answer 10

dont forget to pop out that chain rule of 5x^2+4

Answer 11

where are you getting the first 5x from that is outside the parentheses
?
I am confused on that

Answer 12

Becuase I thought for the first derivative it is just the derivative of:

(5x^2 + 4) ^1/2 ... --> 1/2(5x^2 +4) ^-1/2?

Answer 13

the chain rule: let u = 5x^2 + 4; u' = 2*5x

\[u^{1/2}\to \frac{u'}{2u^(1/2)}\to \frac{\cancel2*5x}{\cancel2\sqrt{5x^2+4}}\]

Answer 14

whenever you have a function wrapped inside of another function, you have to apply the chain rule ... which is hard to grasp at first since they dont tell you that you always pop out a chain rule even when you were learning the process

Answer 15

oh okay I gotcha, so, can we back this up for a moment...

In order to find the first derivative of this problem, y = the square root of 5x^2 + 4....

you put 1/2(5x^2 + 4) ^-1/2 ? and that is the first derivative?

Answer 16

thats incomplete ... so no

Answer 17

So is it 1/2(10x)^-1/2? or am I doing that derivative incorrectly

Answer 18

do it in steps .. you seem to be confusing a few tings at the moment

Answer 19

\[\frac d{dx}[f(x)]^n=n[f(x)]^{n-1}~*~\frac d{dx}f(x)\]

Answer 20

let f(x) = 5x^2 + 4

the derivative of f(x) is therefore 10x

and n=1/2

fill in the parts

\[\frac d{dx}[f(x)]^n=n[f(x)]^{n-1}~*~\frac d{dx}f(x)\]

\[\frac d{dx}[5x^2+4]^{1/2}=\frac12[5x^2+4]^{-1/2}~*~10x\]

Answer 21

So, do we need to simplify that for the first derivative to be complete?

Answer 22

you can if you want .... 10/2 = 5 is what i did up top

Answer 23

\[5x~(5x^2+4)^{-1/2}\]
now this is a product rule ... and the chain rule will still apply for this one too
\[5'x~(5x^2+4)^{-1/2}~+~5x~(5x^2+4)'^{-1/2}\]
\[5~(5x^2+4)^{-1/2}~-~5x~\frac12(5x^2+4)'^{-3/2}*10x\]
\[5~(5x^2+4)^{-1/2}~-~25x(5x^2+4)^{-3/2}\]
factor out the -3/2 stuff
\[(5~(5x^2+4)^{4/2}~-~25x)(5x^2+4)^{-3/2}\]
\[\frac{5~(5x^2+4)^{4/2}~-~25x}{(5x^2+4)^{3/2}}\]
and the rest is simple algebra stuff

Answer 24

mighta factored wrong up top :) make that a 1/2 exponent instead

Answer 25

\[\frac{5~(5x^2+4)^{1/2}~-~25x}{(5x^2+4)^{3/2}}\]
thats feels better